Hw4 Problem 4

### Problem 4

For a fixed point $(a,b) \in \mathbb{R^{2}}$, define $T_{a,b} : \mathbb{R^{2}} \rightarrow \mathbb{R^{2}}$ by $T_{a,b}(x,y) = (a + x, b + y)$. Show that $G = \{ T_{a,b} | a,b \in \mathbb{R} \}$ is a group under function composition $\circ$.

Solution

$\mathscr{G_{1}}$: Let $a,b,c,d,e,f \in \mathbb{R}$. Now,

(1)
\begin{align} ((T_{a,b} \circ T_{c,d} ) \circ T_{e,f}) (x,y) = (T_{a,b}\circ T_{c,d})(x+e, y+f) \end{align}
(2)
$$= T_{a,b}((T_{c,d}(x+e, y+f))$$
(3)
$$= T_{a,b}(x+e+c, y+f+d)$$
(4)
$$= (x+e+c+a, y+f+d+b)$$

Similarly,

(5)
\begin{align} T_{a,b}\circ(T_{c,d}\circ T_{e,f})(x,y) = T_{a,b}(T_{c,d}(T_{e,f}(x,y))) \end{align}
(6)
$$= (x+e+c+a, y+f+d+b)$$

Hence, $T_{a,b}\circ (T_{c,d}\circ T_{e,f}) = (T_{a,b}\circ T_{c,d})\circ T_{e,f}$ and the operation is associative.

$\mathscr{G_{2}}$: Let $a,b \in \mathbb{R}$, and clearly $0 \in \mathbb{R}$. Now,

(7)
\begin{align} (T_{a,b}\circ T_{0,0})(x,y) = T_{a,b}(x+0, y+0) \end{align}
(8)
$$= (a + x, b + y)$$
(9)
$$= T_{a,b}(x,y)$$
(10)
\begin{align} = (T_{0,0} \circ T_{a,b})(x,y) \end{align}

Hence $e = T_{0,0}$, and there is an identity.

$\mathscr{G_{3}}$: Let $a,b, -a, -b \in \mathbb{R}$. Now,

(11)
\begin{align} (T_{a,b} \circ T_{-a,-b})(x,y) = T_{a,b}(x-a, y-b) \end{align}
(12)
$$= (x-a+a, y-b+b)$$
(13)
$$= (x,y)$$
(14)
\begin{align} = (T_{-a, -b} \circ T_{a,b})(x,y) \end{align}
(15)
$$= T_{0,0}(x,y)$$

Hence, for any element $T_{a,b}\in G$, $G$ contains an inverse $T_{-a,-b}$

Therefore, $G = \{ T_{a,b} | a,b \in \mathbb{R} \}$ is a group under function composition $\circ$.